Limit of H(x) for between graphs

[furi0n]

Homework Statement

if the graph pf y=h(x) always lies between the graphs of y=x^3 and Y=x^1/3 for what real numbers a can you determine the value of limH(x)x->a? explain and find the limit for each of these values of a

Homework Equations

The Attempt at a Solution

i can draw two graphs but i cannot understand the graph of h(x) there are a lot of real numbers between these two graphs which one should i use

Not: please only help mi finding out values of a .. :)

 

[fzero]

What can you say about h(x) near the point(s) where x³ and x^1/3 approach one another?

 

[furi0n]

excuse me ? does this mean points between [-1,1] interval?

 

[fzero]

excuse me ? does this mean points between [-1,1] interval?

What I mean is that you said you sketched the graphs. Where do the graphs of x³ and x^1/3 meet? Is there anything that you can say about the value of h(x) at those points? Can you say anything about the value of h(x) as x\rightarrow \pm \infty?

 

[furi0n]

What I mean is that you said you sketched the graphs. Where do the graphs of x³ and x^1/3 meet? Is there anything that you can say about the value of h(x) at those points? Can you say anything about the value of h(x) as x\rightarrow \pm \infty?

The graphs meet (1,1) and (-1,-1), no there isn't because i don't know anything about h(x). Then İ can't find any limit of numbers, actually ı don't understand what we did. i haven't understand Question yet

 

[HallsofIvy]

Yes, you don't understand the question! If x^3&lt; h(x)&lt; x^{1/3} (which happens for x< -1 and 0< x< 1) then you know that \lim_{x\to a}x^3&lt; \lim_{x\to a} h(x)\le \lim_{x\to a} x^{1/3} which, since x^3 and x^{1/3} are continuous, is the same as a^3\le \lim_{x\to a} h(x)\le a^{1/3}.<br /> <br /> If x^{1/3}&amp;lt; h(x)&amp;lt; x^3 which happens for -1&lt; x&lt; 0 or x&gt; 1, then, similarly, a^{1/3}\le h(x)\le a^3.<br /> <br /> In general, \lim_{x\to a} h(x) could be <b>any</b> number between those bounds. But what happens when a^3= a^{1/3}? For what a does that happen?

 

[furi0n]

Yes, you don't understand the question! If x^3&lt; h(x)&lt; x^{1/3} (which happens for x< -1 and 0< x< 1) then you know that \lim_{x\to a}x^3&lt; \lim_{x\to a} h(x)\le \lim_{x\to a} x^{1/3} which, since x^3 and x^{1/3} are continuous, is the same as a^3\le \lim_{x\to a} h(x)\le a^{1/3}.<br /> <br /> If x^{1/3}&amp;lt; h(x)&amp;lt; x^3 which happens for -1&lt; x&lt; 0 or x&gt; 1, then, similarly, a^{1/3}\le h(x)\le a^3.<br /> <br /> In general, \lim_{x\to a} h(x) could be <b>any</b> number between those bounds. But what happens when a^3= a^{1/3}? For what a does that happen?

<br /> <br /> Okey , now i understood then x=1,-1 and 0 limh(x)= 1,-1,0 we learned this theorem today. now i understand this perfectly thanks a lot